Wednesday 11 September 2019

Finding the Christoffel symbols and going round in circles

In section 3.3 Carroll proves that the geodesic equation gives a line that parallel transports vectors between two points. It's 'embarrassing simple'. He also claims that this line is the shortest distance between the two points. The geodesic equation is
\frac{d^2x^\sigma}{d\lambda^2}+\Gamma_{\mu\nu}^\sigma\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}=0&\phantom {10000}(1)\nonumber
\end{align}and the Christoffel symbols are given by
\Gamma_{\mu\nu}^\sigma=\frac{1}{2}g^{\sigma\rho}\left(\partial_\mu g_{\nu\rho}+\partial_\nu g_{\rho\mu}-\partial_\rho g_{\mu\nu}\right)&\phantom {10000}(2)\nonumber
\end{align}He then uses calculus of variations to find the shortest distance and, low and behold, it produces the geodesic equation. So now we know that the geodesic equation which started by just parallel transporting also gives a straight line.

But he then says that it is actually easier to find the Christoffel symbols by using calculus of variations and then disentangling them from the result than by using (2) and he tells us that we will see this in section 3.5. In contrast Orodruin had said that it is better to find the geodesic equation by writing down the Euler-Lagrange equations for the shortest path. I don't know whether I'm coming or going! It is twisted.

Now I've been through the pain of the first part of section 3.5: finding the Christoffel symbols. I followed Carroll carefully and it involves:
  • doing the variation, 
  • integration by parts, 
  • eliminating boundary terms 
  • and then identifying what's left with the geodesic equation and the Christoffel symbols therein. 
Everything except the last part is almost a re-enactment of the proof of the Euler-Lagrange equation. The last part seems like work of the devil.

It seems much easier to find the Christoffel symbols directly from (2) below as we shall see. Since the metric is often diagonal almost all the terms in (2) vanish. It is also very important to remember that for any function ## f## where ## \mu\neq\nu##
\partial_\mu\left(f\left(x^\nu\right)\right)=\frac{\partial}{\partial x^\mu}\left(f\left(x^\nu\right)\right)=0&\phantom {10000}(3)\nonumber
\end{align}as I forgot to my shame many moons ago. (The rule applies if other coordinates are part of the function as long as none of them are the ##x^\mu## coordinate.) So even more terms vanish from (2).

Full details in here

The geodesic equation can also be proved by finding the shortest distance (proper time in GR) between two points (events).
Commentary 3.3 Shortest path by calculus of variations.pdf (3 pages)

Calculating Christoffel symbols by both methods for a simple expanding universe
Commentary 3.5 Expanding Universe Revisited #1.pdf

No comments:

Post a Comment